%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% exercise set 4 - solutions %%% september 30, 2009 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass{amsart} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \usepackage{nicefrac} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} % \input hw.sty \textwidth = 6 in \textheight = 8.5 in \oddsidemargin = 0.0 in \evensidemargin = 0.0 in \topmargin = 0.0 in \headheight = 0.0 in \headsep = 0.0 in \parskip = 6pt \parindent = 0.0in % \hoffset=-.7in % \voffset=-.7in \newcommand{\ds}{\displaystyle} \newcommand{\mA}{{\mathbb A}} \newcommand{\mB}{{\mathbb B}} \newcommand{\mD}{{\mathbb D}} \newcommand{\mN}{{\mathbb N}} \newcommand{\mQ}{{\mathbb Q}} \newcommand{\mR}{{\mathbb R}} \newcommand{\mS}{{\mathbb S}} \newcommand{\mZ}{{\mathbb Z}} \newcommand{\cA}{{\mathcal A}} \newcommand{\cB}{{\mathcal B}} \newcommand{\cC}{{\mathcal C}} \newcommand{\cE}{{\mathcal E}} \newcommand{\cI}{{\mathcal I}} \newcommand{\cP}{{\mathcal P}} \newcommand{\cT}{{\mathcal T}} \newcommand{\cU}{{\mathcal U}} \begin{document} \begin{center} Math 445, Fall 2009 \hfill Exercise Set \#4 \hfill Solutions \end{center} \bigskip {\bf 1.} [\#1, page 78] ~ Let $M$ be a metric space with metric $D$. Prove that if $\{x_n \mid n =1,2, \ldots \} \subset M$ is a sequence which converges to points $x \in M$ and $y \in M$, then $x = y$. \proof Given $\epsilon > 0$ we will show that $D(x,y) < \epsilon$. Since $D(x,y) \geq 0$, and $D(x,y) > 0$ would contradict $D(x,y) < \epsilon$ for $\epsilon = D(x,y)$, we must have $D(x,y) = 0$, so $x = y$. Since $x_n \to x$ there is some integer $N_{\epsilon}$ so that $n \geq N_{\epsilon}$ implies $D(x_n, x) < \epsilon/2$. Since $x_n \to y$ there is some integer $M_{\epsilon}$ so that $n \geq M_{\epsilon}$ implies $D(x_n, y) < \epsilon/2$. Let $n \geq \max\{N_{\epsilon} , M_{\epsilon}\}$. Then $D(x,y) \leq D(x, x_n) + D(x_n, y) < \epsilon/2 + \epsilon/2 = \epsilon$ \endproof \medskip {\bf 2.} [\#4, page 78] ~ Given distinct points $x$ and $y$ in a metric space $M$, prove that there exist open sets $U$ and $V$ such that $x \in U$, $y \in V$, and their \underline{closures} $\overline{U} \cap \overline{V} = \emptyset$. \proof Let $R = D(x,y)$ and set $\epsilon = R/4$. Let $U = B(x,\epsilon)$ and $V= B(y,\epsilon)$. The closed ball $\{ z \in M \mid D(x,z) \leq \epsilon\}$ is closed, and contains $B(x,\epsilon)$, so its closure satisfies $\overline{B(x,\epsilon)} \subset \{ z \in M \mid D(x,z) \leq \epsilon\}$. [They do not have to be equal - give an example.] Likewise, the closure $\overline{B(y,\epsilon)} \subset \{ z \in M \mid D(y,z) \leq \epsilon\}$. Then $$\overline{B(x,\epsilon)} \cap \overline{B(y,\epsilon)} ~ \subset ~ \{ z \in M \mid D(x,z) \leq \epsilon\} \cap \{ z \in M \mid D(y,z) \leq \epsilon\} ~ = ~ \emptyset$$ where the two closed balls are disjoint by the Triangle Inequality. \endproof \medskip {\bf 3.} [\#8, page 79] ~ Let $M$ be a metric space with metric $D$. Prove that the diameter of a set $A$ in $M$ equals the diameter of its closure, $\overline{A}$. \proof The diameter of the sets $A$ and $\overline{A}$ are \begin{eqnarray*} {\rm diam}(A) & = & \sup \{D(x,y) \mid x,y \in A\}\\ {\rm diam}(\overline{A}) & = & \sup \{D(x,y) \mid x,y \in \overline{A}\} \end{eqnarray*} Since $A \subset \overline{A}$ the inequality ${\rm diam}(A) \leq {\rm diam}(\overline{A}) $ is immediate. Suppose that ${\rm diam}(A) < {\rm diam}(\overline{A})$ holds. Choose $\epsilon$ so that $0 < \epsilon < {\rm diam}(\overline{A}) - {\rm diam}(A)$. Now use the ``three epsilon trick''. By definition of the supremum, there exists $u,v \in \overline{A}$ with $D(u,v) > {\rm diam}(\overline{A}) - \epsilon/3$. This is the same as ${\rm diam}(\overline{A}) < D(u,v) + \epsilon/3$. Then $u \in \overline{A}$ implies there exists $u_* \in A \cap B(u,\epsilon/3)$. Likewise, $v \in \overline{A}$ implies there exists $v_* \in A \cap B(v,\epsilon/3)$. By the Triangle Inequality, $${\rm diam}(\overline{A}) ~ < ~ D(u,v) + \epsilon/3 ~ < ~ \epsilon/3 + D(u_*, v_*) + \epsilon/3 + \epsilon/3 ~ \leq ~ {\rm diam}(A) + \epsilon < {\rm diam}(\overline{A})$$ which is a contradiction. \endproof {\bf 4.} [\#11*, page 79] ~ Prove that in a metric space, the closure of a \underline{countable set} has cardinal number at most $c$. [Recall that $c$ is the cardinal of the continuum $\mR$, which equals the cardinal of the power set of the natural numbers, ${\mathcal P}(\mN)$.] \proof For each $u \in \overline{A}$, there exists a convergent sequence $\{x_n \mid n =1,2, \ldots \} \subset A$ with $x_n \to u$. By Problem \#1) above, a convergent sequence $\{y_n\} \subset A$ has a unique limit point, so the number of points of $\overline{A}$ is at most the cardinality of the set of convergent sequences in $A$. The set of convergent sequences $\{x_n\} \subset A$ is a subset of the set of all sequences in $A$. A sequence $\{x_n\}$ can be viewed as a function $F_x \colon \mN \to A$, where $F_x(n) = x_n \in A$. The graph of a function $F \colon \mN \to A$ is a subset of the product $\mN \times A$. Thus, the set of all sequences has cardinality bounded above by the cardinality of the power set $\cP(\mN \times A)$, which has the cardinality $c$ of the continuum. (Both $A$ and $\mN$ are countable, and $\mN$ is countably infinite.) Combining these statements, we have that the cardinality of $\overline{A}$ is bounded above by the cardinality of $\cP(\mN \times A)$; that is, there are at most $c$ points which are the limits of such sequences. \endproof \medskip {\bf 5.} [\#12*, page 79] ~ Prove that the following statements are equivalent for a metric space $M$: \begin{enumerate} \item[(a)] Every subset of $M$ is either open or closed; \item[(b)] At most one point of $M$ is not isolated. \end{enumerate} \proof We first prove that $(b)$ implies $(a)$. If every point of $M$ is isolated, then every point is open, so every subset is a union of open sets, hence is open. [Since every point in a metric space is also closed, this means there are subsets which are both open and closed.] Next, suppose there is one non-isolated point, say $u \in M$. If $A \subset M$ with $u \not\in A$ then $A$ is a union of isolated points, so is open. If $A \subset M$ with $u \in A$ then we claim $A = \overline{A}$ is its own closure, hence $A$ is closed. This follows, since the closure of $A$ is the union of $A$ with its limit points. A limit point is not isolated, so $u$ is the only possible limit point in $M$. As $u \in A$, it contains all of its limit points, so $A = \overline{A}$. Next, show the converse, that $(a)$ implies $(b)$. We show this by contradiction. Suppose that $M$ has more than one non-isolated point, say $u \ne v \in M$. Choose a sequence $\{a_n \mid n = 1,2, \ldots\} \subset M - \{u,v\}$ with $a_n \to u$. We can also assume that each $a_n$ satisfies $D(u,a_n) \leq D(u,v)/2$. Set $A = \{a_n \mid n = 1,2, \ldots\} \cup \{v\}$. Then $A$ is not closed, as $u$ is a limit point of $A$, but $u \not\in A$. Also, $A$ is not open, as every open ball $B(v,\epsilon) \cap M$ contains a point other than $v$, since $v$ is a limit point. For all $\epsilon < D(u,v)/2$ we have $B(v,\epsilon) \cap A = \emptyset$. So, $B(v,\epsilon) \not\subset A$. Thus $A$ is not open. \endproof \bigskip {\bf 6.} [\#13*, page 79] ~ Let $M$ be a metric space in which the closure of every open set is open. Prove that $M$ is discrete. That is, show that every point of $M$ is an open set. \proof First, this condition is very strange, as the closure of an open set $U$ is closed! How can it be open? The closure $\overline{U}$ is also open means its complement is closed, which means that every convergent sequence $\{z_n\}$ in the complement of $\overline{U}$ with $z_n \to z$, then $z$ must also lie in the complement. We give a proof by contradiction. Assume that $M$ is not discrete, so there exists a non-isolated point $u \in M$. We will construct an open set $U$ as a union of open balls, such that the closure $\overline{U}$ contains a limit point of the complement of $\overline{U}$. This will show the complement of $\overline{U}$ is not closed. [Note - the construction below gets complicated, so it is likely there is a simpler example.] Inductively choose a sequence $A = \{a_n \mid n = 1,2, \ldots\} \subset M - \{u\}$ such that $a_n \to u$, and so that $D(u,a_{n}) < \frac{1}{2} D(u, a_{n-1})$. That is, $a_{n}$ is less than half the distance to $u$ than the previous point $a_{n-1}$. Since a convergent sequence has a unique limit point by Problem \#1) the set $A$ has a unique limit point $u$, hence $\overline{A} = A \cup \{u\}$. Set $\epsilon_{n} = \frac{1}{2} D(u, a_{n})$. Then by choice, $$D(u,a_{n+1}) < \frac{1}{2} D(u, a_{n}) = \epsilon_n \quad \Longrightarrow \quad \epsilon_{n+1} < \frac{1}{2} \epsilon_n$$ Also, $D(u,a_{n+1}) < \frac{1}{2} D(u, a_{n})$ implies that $D(a_n ,a_{n+1}) > \frac{1}{2} D(u, a_{n}) = \epsilon_n$ by the Triangle Inequality. Since $\epsilon_n < \epsilon_{n-1}$ we also have that $D(a_{n-1} ,a_{n}) > \epsilon_{n-1} > \epsilon_n$. Combining these estimates we get that $$B(a_n, \epsilon_n/2) \cap A = \{a_n\} \quad \& \quad B(a_m, \epsilon_m/2) \cap B(a_n, \epsilon_n/2) \quad \text{ for all } m \ne n \geq 1 $$ In other words, by choosing the sequence $\{a_n\}$ to converge to $u$ exponentially fast, we are able to construct a sequence of open metric balls about the points of the sequence which are all disjoint. Consider the set of points in the sequence with even index, $B = \{a_{2n} \mid n = 1, 2, 3, \ldots\}$. Then $u$ is also a limit point of $B$. Define the open set $U$ to be the union of the open balls around these points with even index: $$U = B(a_2 , \epsilon_2/2) \cup B(a_4, \epsilon_4/2) \cup \cdots \cup B(a_{2n}, \epsilon_{2n}/2) \cup \cdots$$ The claim is that the closure $\overline{U}$ is not open, or that the complement $M - \overline{U}$ contains a convergent sequence whose limit lies in $\overline{U}$. First, note that $u \in \overline{U}$ as $u \in \overline{B}$. Also, $u$ is the limit of the odd terms in the sequence $\{a_n\}$ as well, so we just need to show that $\{a_{2n-1} \mid n = 1, 2, 3, \ldots\} \cap \overline{U} = \emptyset$. Suppose that there exists a sequence $\{b_{\ell}\} \subset U$ with $b_{\ell} \to a_{m}$. We claim that $m$ must be even. As the sequence converges, there exists $\ell_0 $ so that $\{b_{\ell} \mid \ell \geq \ell_0\} \subset B(a_m, \epsilon_m/2)$. But $\{b_{\ell}\} \subset U$ and $U \cap B(a_n, \epsilon_n/2) = \emptyset$ for $n$ odd implies that $m$ is even. It follows that no odd term $a_{2n-1}$ is a limit point of $U$. Hence $\overline{U}$ cannot be open. \endproof \bigskip {\bf 7.} [\#14*, page 79] ~ Prove that in a metric space, every open set is the union of a countable number of closed sets. Deduce from this that every closed set is the intersection of a countable number of open sets. \proof The second part follows from the first by de~Morgan's laws. Let $U \subset M$ be an open set, and $F = M - U$ its complement which is closed. Notice that $U$ open implies that for each $x \in U$ choose $\epsilon_x > 0$ such that $B(x, \epsilon_x) \subset U$. Then for any point $y \in F$, the distance $D(x,y) \geq \epsilon_x$ and so $D(x, F) \geq \epsilon_x$. For each $n \geq 1$, define $$V_n = \bigcup_{y \in F} ~ B(y, 1/n)$$ Then each $V_n$ is open as it is a union of open sets, and $F \subset V_n$ for all $n$. Finally, note that given any $x \in U$, if $n$ satisfies $1/n < \epsilon_x$ then $x \not\in V_n$. Let $K_n = M - V_n$ be the complement of $V_n$. Each $F_n$ a closed subset. Since $F \subset V_n$ we have $K_n \subset U$ for each $n$. Finally, for each $x \in U$ if $n > 1/\epsilon_x$ then $x \in K_n$. Thus, each every point of $U$ is contained in some $K_n$ hence $\ds U ~ = ~ \bigcup_{n=1}^{\infty} ~ K_n$. \endproof \vfill \eject {\bf 8.} [\#16, page 79] ~ Prove that a metric space is discrete if and only if every convergent sequence is ultimately constant. \proof The metric space $M$ is discrete means, by definition, that every subset of $M$ is open. In particular, for each point $y \in M$ the singleton set $\{y\}$ is open. Now suppose that $x_n \to x$ is a convergent sequence. Then $\{x\}$ is open implies there is some $\epsilon > 0$ such that $B(x,\epsilon) \subset \{x\}$. That is, $\{y \in M \mid D(x,y) < \epsilon\} = \{x\}$. [This means that there is no point of $M$ within distance $\epsilon$ of $x$.] Now $x_n \to x$ means there exists $N_{\epsilon}$ an integer such that $n \geq N_{\epsilon}$ implies $D(x, x_n) < \epsilon$. But $x_n \in B(x,\epsilon)$ implies $x_n = x$ as the open ball consist of the singleton, $\{x\}$. Conversely, suppose that $M$ is not discrete, then there is some point $x \in X$ which is not itself open. That is, $x$ is not isolated. By the Limit Lemma, there exists a sequence $\{x_n\} \subset M$ of distinct points with $x_n \to x$ and $\epsilon_n = D(x, x_n) > 0$ so that $\epsilon_n \to 0$ monotonically. Thus, $n \ne m$ implies $x_n \ne x_m$ as their distances from $x$ differ, so the sequence $\{x_n\}$ is not ultimately constant. \endproof \bigskip {\bf 9.} [\#17, page 79] ~ Prove that a metric space is discrete if and only if it has no limit points. \proof The metric space $M$ is discrete means, by definition, that every subset of $M$ is open. In particular, each point $x \in M$ defines an open set $\{x\}$. Then $\{x\}$ is open implies there is some $\epsilon > 0$ such that $B(x,\epsilon) \subset \{x\}$. That is, $U = \{y \in M \mid D(x,y) < \epsilon\} = \{x\}$. Thus, $x$ is isolated. Conversely, suppose that there exists $x \in M$ and a sequence $\{x_n\} \subset M - \{x\}$ with $x_n \to x$. Then for every open set $U$ with $x \in U$ the intersection $U \cap \{x_n\}$ contains an infinite number of points, hence $x$ is not isolated. That is, $\{x\}$ is not an open set. \endproof \bigskip {\bf 10.} [\#19*, page 79] ~ If a metric space $M$ has only countably many open sets, prove that $M$ is finite. \proof Suppose that $M$ is not finite, then we show there exists uncountably many distinct open subsets. First, if $M$ is discrete, then every subset $A \in \cP(M)$ is open, and $M$ infinite implies the cardinality of $\cP(M)$ is uncountable. If $M$ is not discrete, then there exists a limit point $x \in M$. Then by the Limit Lemma, there is a sequence of points $\{x_n\} \subset M - \{x\}$ with $\epsilon_n = D(x,x_n) > 0$ and $\epsilon_1 > \epsilon_2 > \ldots > \epsilon_n \to 0$. Set $\delta_n = \min\{(\epsilon_n - \epsilon_{n+1})/2 , (\epsilon_{n-1} - \epsilon_{n})/2 \}$. Then by the triangle inequality, for each $n \geq 1$, $$B(x_n, \delta_n) ~ \cap ~ B(x_{n+1}, \delta_{n+1}) = \emptyset $$ This produces a countably infinite collection $\cU = \{B(x_n, \delta_n) \mid n =1 , 2, \ldots\}$ of open sets. Each subset of $\cU$ is a collection of open balls, so their union if an open subset of $M$. Distinct subsets give rise to distinct open sets in $M$, as the open balls are disjoint. The cardinality of $\cP(\cU)$ is uncountable, so this produces an uncountable collection of distinct open subsets of $M$. \endproof \vfill \end{document}